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Interaction-picture and gell-mann low

  1. Dec 1, 2014 #1
    Suppose you want to evaluate:

    $$\langle \Omega | T\phi(x) \phi(y) | \Omega \rangle$$

    where [itex]\Omega [/itex] is the ground state of the full Hamiltonian, and the fields are in the Heisenberg representation. Assume [itex]x_0 > y_0 [/itex] for now. Then it's straightforward to show that

    $$\langle \Omega | T\phi(x) \phi(y) | \Omega \rangle=\langle \Omega | S^\dagger(t,0)\phi_I(x) S(x_0, y_0) \phi_I(y) S(y_0,0) | \Omega \rangle$$

    where the fields are in the interaction representation [itex] \phi_I(x)=\phi_F(x)[/itex], [itex]S(t,t')=U(t)U^\dagger(t') [/itex], and [itex]U(t)=e^{iH_0t}e^{-iHt}[/itex], where subscript [itex]F [/itex] stands for the free-field.

    According to Wikipedia, the Gell-Mann Low formula is:

    $$ |\Psi^{(\pm)}_\epsilon \rangle = \frac{ U_{\epsilon I} (0,\pm\infty) |\Psi_0 \rangle}{\langle \Psi_0 | U_{\epsilon I}(0,\pm\infty)|\Psi_0\rangle} $$

    I want to use this in my equation to have an expression in terms of the noninteracting vacuum [itex]|\Psi_0\rangle [/itex] instead of the interacting one [itex]|\Omega \rangle[/itex].

    Is it correct to say that [itex]|\Psi^{(-)}_\epsilon \rangle [/itex] is equal to my [itex]|\Omega \rangle [/itex], and [itex]\langle \Psi^{(+)}_\epsilon | [/itex] is equal to my [itex]\langle \Omega| [/itex]?

    If I do this, then I get:

    $$\langle \Omega | T\phi(x) \phi(y) | \Omega \rangle=\frac{\langle \Psi_0 |U^\dagger_{\epsilon I} (0, \infty) S^\dagger(x_0,0)\phi_I(x) S(x_0, y_0) \phi_I(y) S(y_0,0) U_{\epsilon I} (0, -\infty) |\Psi_0 \rangle}{\langle \Psi_0 | U_{\epsilon I}(0,\infty)|\Psi_0\rangle^* \langle \Psi_0 | U_{\epsilon I}(0,-\infty)|\Psi_0\rangle}$$

    Now assume that their [itex]U_{\epsilon I}[/itex] corresponds to my [itex]S [/itex]. Then using the property that [itex]S^\dagger(t,t')=S(t',t) [/itex] and [itex]S(t,t')S(t',t'')=S(t,t'') [/itex] I get:

    $$\langle \Omega | T\phi(x) \phi(y) | \Omega \rangle=\frac{\langle \Psi_0 |S(\infty,x_0)\phi_I(x) S(x_0, y_0) \phi_I(y) S(y_0, -\infty) |\Psi_0 \rangle}{\langle \Psi_0 | S(\infty,0)|\Psi_0\rangle \langle \Psi_0 | S(0,-\infty)|\Psi_0\rangle}$$

    This is almost the Green's function written in the interaction picture, except the denominator is wrong. The correct expression is:

    $$\langle \Omega | T\phi(x) \phi(y) | \Omega \rangle=\frac{\langle \Psi_0 |TS(\infty,-\infty)\phi_I(x) \phi_I(y) |\Psi_0 \rangle}{\langle \Psi_0 | S(\infty,-\infty)||\Psi_0\rangle}$$

    I don't think the two denominators are equal. In general you need a complete set of unperturbed energy eigenstates, and not just one, to make the two denominators equal.

    Also, what is the meaning of choosing two different vacuums for the bra and ket, [itex]|\Psi^{(\pm)}_\epsilon \rangle [/itex], for the interacting ground state? I thought we usually assume the interacting ground state is nondegenerate? But it seems that it's doubly degenerate.
  2. jcsd
  3. Dec 7, 2014 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
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